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Leo Lam © 2010-2011 Signals and Systems EE235
Leo Lam © 2010-2011 Breeding What do you get when you cross an elephant and a zebra? Elephant zebra sin theta
Leo Lam © 2010-2011 Today’s menu Friday: LCCDE Zero-Input Response Today: LCCDE Zero-State Response (forced)
Zero input response (example) Leo Lam © 2010-2011 4 4 steps to solving Differential Equations: Step 1. Find the zero-input response = natural response y n (t) Step 2. Find the Particular Solution y p (t) Step 3. Combine the two Step 4. Determine the unknown constants using initial conditions
From Friday (example) Leo Lam © 2010-2011 5 Solve Guess solution: Substitute: We found: Solution: Characteristic roots = natural frequencies/ eigenvalues Unknown constants: Need initial conditions
Zero-state output of LTI system Leo Lam © 2010-2011 6 Response to our input x(t) LTI system: characterize the zero-state with h(t) Initial conditions are zero (characterizing zero-state output) Zero-state output: Total response(t)=Zero-input response (t)+Zero-state output(t) T (t) h(t)
Zero-state output of LTI system Leo Lam © 2010-2011 7 Zero-input response: Zero-state output: Total response: Total response(t)=Zero-input response (t)+Zero-state output(t) “Zero-state”: (t) is an input only at t=0 Also called: Particular Solution (PS) or Forced Solution
Zero-state output of LTI system Leo Lam © 2010-2011 8 Finding zero-state output (Particular Solution) Solve: Or: Guess and check Guess based on x(t)
Trial solutions for Particular Solutions Leo Lam © 2010-2011 9 Guess based on x(t) Input signal for time t> 0 x(t) Guess for the particular function y P
Particular Solution (example) Leo Lam © 2010-2011 10 Find the PS (All initial conditions = 0): Looking at the table: Guess: Its derivatives:
Particular Solution (example) Leo Lam © 2010-2011 11 Substitute with its derivatives: Compare:
Particular Solution (example) Leo Lam © 2010-2011 12 From We get: And so:
Particular Solution (example) Leo Lam © 2010-2011 13 Note this PS does not satisfy the initial conditions! Not 0!
Natural Response (doing it backwards) Leo Lam © 2010-2011 14 Guess: Characteristic equation: Therefore:
Complete solution (example) Leo Lam © 2010-2011 15 We have Complete Sol n : Derivative:
Complete solution (example) Leo Lam © 2010-2011 16 Last step: Find C 1 and C 2 Complete Sol n : Derivative: Subtituting: Two equations Two unknowns
Complete solution (example) Leo Lam © 2010-2011 17 Last step: Find C 1 and C 2 Solving: Subtitute back: Then add u(t): y n ( t ) y p ( t ) y ( t )
Another example Leo Lam © 2010-2011 18 Solve: Homogeneous equation for natural response: Characteristic Equation: Therefore: Input x(t)
Another example Leo Lam © 2010-2011 19 Solve: Particular Solution for Table lookup: Subtituting: Solving: b=-1, =-2 No change in frequency! Input signal for time t> 0 x(t) Guess for the particular function y P
Another example Leo Lam © 2010-2011 20 Solve: Total response: Solve for C with given initial condition y(0)=3 Tada!
Stability for LCCDE Leo Lam © 2010-2011 21 Stable with all Re( j <0 Given: A negative means decaying exponentials Characteristic modes
Stability for LCCDE Leo Lam © 2010-2011 22 Graphically Stable with all Re( j )<0 “Marginally Stable” if Re( j )=0 IAOW: BIBO Stable iff Re(eigenvalues)<0 Im Re Roots over here are stable
Leo Lam © 2010-2011 Summary Differential equation as LTI system Stability of LCCDE
Leo Lam © Signals and Systems EE235 Lecture 19.
Leo Lam © Signals and Systems EE235 Leo Lam © Stanford The Stanford Linear Accelerator Center was known as SLAC, until the big earthquake,
WEEK 1 You have 10 seconds to name…
Addition 1’s to
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TWO STEP EQUATIONS 1. SOLVE FOR X 2. DO THE ADDITION STEP FIRST
Leo Lam © Signals and Systems EE235. Leo Lam © Today’s menu Yesterday: Exponentials Today: Linear, Constant-Coefficient Differential.
SUBTRACTING INTEGERS 1. CHANGE THE SUBTRACTION SIGN TO ADDITION 2. TAKE THE INVERSE OF THE SECOND NUMBER 3. FOLLOW THE RULES FOR ADDITION 4. ADD THE OPPOSITE.
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DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.
© S Haughton more than 3?
1 Unit 1 Kinematics Chapter 1 Day
1 Copyright © 2010, Elsevier Inc. All rights Reserved Fig 2.1 Chapter 2.
Leo Lam © Signals and Systems EE235 Leo Lam © Today’s menu Exponential response of LTI system LCCDE Midterm Tuesday next week.
Squares and Square Root WALK. Solve each problem REVIEW:
MULT. INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.
Jeopardy Topic 1Topic Q 1Q 6Q 11Q 16Q 21 Q 2Q 7Q 12Q 17Q 22 Q 3Q 8Q 13Q 18Q 23 Q 4Q 9Q 14Q 19Q 24 Q 5Q 10Q 15Q 20Q 25 Final Jeopardy.
Leo Lam © Signals and Systems EE235 Lecture 18.
Leo Lam © Signals and Systems EE235. Courtesy of Phillip Leo Lam ©
Solve an equation by multiplying by a reciprocal EXAMPLE 5 SOLUTION Write original equation. x = 4x = – x = – Solve 7 2 – The coefficient.
MULTIPLICATION EQUATIONS 1. SOLVE FOR X 3. WHAT EVER YOU DO TO ONE SIDE YOU HAVE TO DO TO THE OTHER 2. DIVIDE BY THE NUMBER IN FRONT OF THE VARIABLE.
We will resume in: 25 Minutes We will resume in: 24 Minutes.
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ADDING INTEGERS 1. POS. + POS. = POS. 2. NEG. + NEG. = NEG. 3. POS. + NEG. OR NEG. + POS. SUBTRACT TAKE SIGN OF BIGGER ABSOLUTE VALUE.
Leo Lam © Signals and Systems EE235. Leo Lam © Happy Tuesday! Q: What is Quayle-o-phobia? A: The fear of the exponential (e).
Simultaneous Equations elimination. What are they? Simply 2 equations –With 2 unknowns –Usually x and y To SOLVE the equations means we find values of.
Chapter 13 How Cells Obtain Energy from Food Essential Cell Biology Third Edition Copyright © Garland Science 2010.
MULTIPLYING MONOMIALS TIMES POLYNOMIALS (DISTRIBUTIVE PROPERTY)
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Copyright © 2011, Elsevier Inc. All rights reserved. Chapter 5 Author: Julia Richards and R. Scott Hawley.
ALGEBRAIC EXPRESSIONS Step 1 Write down the question Step 2 Plug in the numbers Step 3 Use PEMDAS Work down, Show all steps.
Past Tense Probe Past Tense Probe – Practice 1 Past Tense Probe – Practice 2.
Substitution. Take OUT the variable and PUT in the new expression y = 4x + 1 2x – y = 9 y = 4x + 1 2x – y = 9 -2x – 3 = 6x + 1 y = -2x -3 y = 6x + 1 y.
Reading Assignment: Chapter 8 in Electric Circuits, 9 th Ed. by Nilsson 1 Chapter 8 EGR 272 – Circuit Theory II 2 nd -order circuits have 2 independent.
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ALGEBRAIC EXPRESSIONS Step 1Write the problem. Step 2Substitute in the values for the unknown (variable). Step 3Use PEMDAS (remember to go left to right).
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